The lines (x-1/2)=(y+1/-3)=(z+10/8) and (x-4/1)=(y+3/k)=(z+1/7) are coplanar if k =

Question

The lines (x-1/2)=(y+1/-3)=(z+10/8) and (x-4/1)=(y+3/k)=(z+1/7) are coplanar if k = (A) 4 (B) -4 (C) 2 (D) -2

Tardigrade Admin · Accepted Answer

Given lines are (x-1/2)=(y-(-1)/-3)=(z-(-10)/8) and (x-4/1)=(y-(-3)/k)=(z-(-1)/7) These lines are coplanar ∴ |4-1&-3-(-1)&-1-(-10) 2&-3&8 1&k&7|=0 ⇒ 3(- 21 - 8k) + 2(14 - 8) + 9(2k + 3) = 0 ⇒ 6k=-24 ∴ k=-4.

Q. The lines $\frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 10}{8}$
and $\frac{x - 4}{1} = \frac{y + 3}{k} = \frac{z + 1}{7}$ are coplanar if $k =$

$4$

$- 4$

$2$

$- 2$

Q. The lines 2x−1​=−3y+1​=8z+10​ and 1x−4​=ky+3​=7z+1​ are coplanar if k=

4

−4

2

−2

Given lines are 2x−1​=−3y−(−1)​=8z−(−10)​ and 1x−4​=ky−(−3)​=7z−(−1)​ These lines are coplanar ∴∣∣​4−121​−3−(−1)−3k​−1−(−10)87​∣∣​=0 ⇒3(−21−8k)+2(14−8)+9(2k+3)=0 ⇒6k=−24 ∴k=−4.

Q. The lines $\frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 10}{8}$
and $\frac{x - 4}{1} = \frac{y + 3}{k} = \frac{z + 1}{7}$ are coplanar if $k =$

$4$

$- 4$

$2$

$- 2$